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A particle is performing SHM along `x-`axis with amplitude `4.0cm` and time period `1.2s` . What is the minimum time is deci`-` second taken by the particle to move from `x=+2cm `to `x=+4cm` and back again. |
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Answer» Correct Answer - 4 As `x=asin omegat =asin((2pi)/(T))t` so, `t=(T)/(2pi)sin^(-1)((x)/(a)),`where`a=4cm.` At `x=2,` `t=(T)/(2pi)sin^(-1)((2)/(4))=(T)/(2pi)xx(pi)/(6)=(T)/(12)=(1.2)/(12)=(1)/(10)` At `x=4` `t=(T)/(2pi)sin^(-1)((4)/(4))=(T)/(2pi)xx(pi)/(2)=(T)/(4)=(1.2)/(4)=(3)/(10)` so, time taken in going from `x=+2cm` to `x=+4cm` will be `(3)/(10)-(1)/(10)=(2)/(10)s=2` deci second, and same time will be taken for coming back. So, total time taken `=2+2=4` deci second |
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